If `e_(1)` and `e_(2)` are the eccentricities of the hyperbola and its conjugate hyperbola respectively then `(1)/(e_(1)^(2))+(1)/(e_(2)^(2))` is equal to

Options

A. `5/4`
B. `4/5`
C. 1
D. `1/2`

Answer

To solve the problem, we need to find the value of \( \frac{1}{e_1^2} + \frac{1}{e_2^2} \), where \( e_1 \) is the eccentricity of the hyperbola and \( e_2 \) is the eccentricity of its conjugate hyperbola.

### Step-by-Step Solution:

1. **Understand the Eccentricities**:
- For a hyperbola given by the equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the eccentricity \( e_1 \) is given by:
\[ e_1 = \sqrt{1 + \frac{b^2}{a^2}} = \frac{\sqrt{a^2 + b^2}}{a} \]
- For the conjugate hyperbola given by the equation \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \), the eccentricity \( e_2 \) is given by:
\[ e_2 = \sqrt{1 + \frac{a^2}{b^2}} = \frac{\sqrt{a^2 + b^2}}{b} \]

2. **Calculate \( e_1^2 \) and \( e_2^2 \)**:
- Squaring the eccentricities, we have:
\[ e_1^2 = \frac{a^2 + b^2}{a^2} \]
\[ e_2^2 = \frac{a^2 + b^2}{b^2} \]

3. **Find \( \frac{1}{e_1^2} \) and \( \frac{1}{e_2^2} \)**:
- Now we can find \( \frac{1}{e_1^2} \) and \( \frac{1}{e_2^2} \):
\[ \frac{1}{e_1^2} = \frac{a^2}{a^2 + b^2} \]
\[ \frac{1}{e_2^2} = \frac{b^2}{a^2 + b^2} \]

4. **Add the Two Fractions**:
- Now, we can add these two fractions:
\[ \frac{1}{e_1^2} + \frac{1}{e_2^2} = \frac{a^2}{a^2 + b^2} + \frac{b^2}{a^2 + b^2} \]
- Since both fractions have the same denominator, we can combine them:
\[ \frac{1}{e_1^2} + \frac{1}{e_2^2} = \frac{a^2 + b^2}{a^2 + b^2} = 1 \]

### Final Result:
Thus, the value of \( \frac{1}{e_1^2} + \frac{1}{e_2^2} \) is:
\[ \boxed{1} \]